Why $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$ is equal to 1?

Question

$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$

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By maths calculator it results 1. I calculate and results $\sqrt{-\frac{1}{2}}$.

$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$

$\sqrt{\frac{-{(3)}^{{\displaystyle\frac12}\times2}}{2^2}+\frac{1^2}{2^2}}=\sqrt{\frac{-3}4+\frac14}=\sqrt{\frac{-3+1}4}=\sqrt{\frac{-2}4}=\sqrt{-\frac12}$

Enlighten me what went wrong?

Answer 1

Hint:$(-\sqrt{3})^2=(-1)^2(3)^{{\frac{1}{2}}2}=3$

Answer 2

The square of a negative number is positive.

$$\left(-\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}$$

Answer 3

There's something called the "Trivial Inequality" that states $$x^2 \geq 0$$ for all $x\in \mathbb{R}$. So $$\left( -\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.$$

Answer 4

$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}=\sqrt{\frac{3}{4}+{\frac14}}=\sqrt{1}=1.$
Your problem was with not powering the minus in the first expression in numerator.
See the difference between $$(-\sqrt3)^2=3$$ and $$-(\sqrt3)^2=-3.$$ As you can see in the first case the minus is inside the brackets which means that the minus has to be raised to the power (opposite situation we have in the second case).